How do you simplify the expression #(sin^2tcot2^t)/(1-sin^2t)#?

Question

3036 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-21T04:41:49

1

I'm guessing here (let me know if I'm wrong of course) that you meant

#(sin^2(t)*cot^2(t))/(1-sin^2(t)), # not #(sin^2(t)*cot(2^t))/(1-sin^2(t))#

If you meant the first one,

#(sin^2(t)*cot^2(t))/(1-sin^2(t))#

(NOTE: #cot(x) = cos(x)/sin(x)# by definition)

#=(sin^2(t)*(\frac{cos^2(t)}{sin^2(t)}))/(1-sin^2(t))#
#=cos^2(t)/(1-sin^2(t))#

(NOTE: use Pythagorean identity #cos^2(x)+sin^2(x)=1#)

#=cos^2(t)/(cos^2(t))#
#=1#

If not, it's a bit more tricky